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113
Definite
integral
summation
2
,
1
=
1
+
1
+
..
+
1
=
n
n
(n+1)
i
=
H
2
+
...
+
n
=
i
=
1
2
n
(n+1)
(2n+
1)
I
i
=
1424
...
+
n
=
--
i
=
1
-Ri
-
1)
=
1
+
3
+
5
+
...
+
2n
-
1)
=
7
n
=
4
-
,
5
I
proof
:
(2i
-1
=
(i)
-
E
=
2)i)
-
n
=
2
.
A
-n
=
nin
-
n
=
n
!
1
+
2
+
3
+
4
+
5 +
....
=
"
-
z
(Ramanujan
Sum)
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Area
Find
the
area
A
of
the
region
under
the
line
y
=
x+
1
·
above
the
x-axis
·
between
x
=
0
and
x
=
3
Idea
:
we
divide
the
region
into
small
rectangle
pieces
·
upper
bound
·
lower
bound
·
take
limit
y
·
P(3
,
4)
3
p13
,
4)
-+
+
+
-
↓
I
N
*
I
-
-
is
Xi-l
E
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We
take
a
partition
of
10
.
33
by
0
=
Xo
<
X
,
<
...
<Xn+1
<
Xn
=
3
.
To
make
the
computation
easier
,
we
take
Xx
=
P
=
0
...
n
On
each
[Xix
,
Xi]
,
the
lower
bound
is
taken
by
f(xit)
,
i
.
e
.
fixit)
=
f(x)
for
all
xe[Xit
.
Xi]
.
The
approximation
from
below
is
↳
=
E
,
f(xi
-
-
=
E
,
/Xi
+1)
=
E(
+
1)
=
((
:
-)
+
=
i) +
.
n
=
-
-
4
+
3
=
2
+
3
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On
(xix
.
xi]
,
the
upper
bound
is
taken
by
f(xi)
,
i
.
e
.
f(x)
<
fixi)
for
all
x=(Xit
,
xi]
.
Un
=
fIXi)
-
=
Xit
=
E
,
(2
+
1)
=
i)
+
E
=
.
int
+
2
.
n
=
=
.
+
3
r
L
<
Area
<
Un
i
I
=
+
3
>
Area
-I
+
3
we
take
the
limit
+
to
,
lim
L
Area
<
lim
Un
n
+
i
n
+
-
Area
=
9 +
3
.
I
1
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In
general
,
for
a
continuous
function
y
=
fix)
on
[a
,
b]
,
·
we
take
partition
P
a
=
Xo<
X
,
<
...
<Xn
+
Xn
=
b
Denote
&xi
=
Xi-Xi-1
·
On
each
piece
[xit
,
Xi]
there
is
a
minimal
point
:
and
a
maximal
point
U
;
10
flil<fix)
<f(Ui)
0x=/Xix
.
Xi)
We
define
217
,
P)
=
=
,
fli)
ox
:
lower
sum
01
,
>
=
E
,
finisox
:
upper
sum
RC
,
P
.
=
E
,
flci)ox
:
Riemann
sum
Ci
=
SXit
,
Xi]
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Then
we
alway
have
<
.p
)
=
R1t
,
P
.
c)
=
Ult
.
P)
.
Let
hips
be
the
number
of
partitions
def
.
and
1P11
=
max
/coxi
1
i
<n(p)
Def
.
Let
f(x)
be
a
function
on
interval
(a
,
b)
If
lim
Lif
.
p)
=
lim
UC
.
P)
for
partition
n(p)
->
A
n(p)
-
0
11P11
->
0
11
P11
-
0
-
P
of
(a
.
b]
with
conditions
n(p)
->
+0
and
/P1-0
,
then
we
say
that
fixe
is
integrable
on
[a
,
33
,
and
we
denote
the
limit
by
I
,
"fixdx
It
follows
that
Safixdx=
lim
RA
.
P
.
23
.
n(p)
+
Il
P11
-
0
-
Theorem
If
f
is
continuous
on
[a
,
b]
then
f
is
integrable
on
[a
.
b3
.
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More
on
the
symbol
/fixdx
I
It
is
a
number
!
10
J
is
the
integral
sign
,
it
is
-many
sum
-
I
2
We
make
convention
that
I
=- 1
!
30
The
function
o
is
called
the
integrand
4
X
is
the
variable
of
the
integrand
5
Ex
is
the
differential
of
X
00
S
,
"
fixdx=
1
?
fits
&t
dummy
variable
7
we
also
write
I
fixsdox
as
Sidxfixs
8
:
In
general
,
we
will
Not
use
above
definition
for
the
computation
.
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Theorem
3
(Properties
of
definite
integral)
Let
f
and
9
be
integrable
on
an
interval
containing
a
,
b
and
c
las
1
,
fixidx
=
0
163
S
,
"
fixsdx
=
-1
,
fix)dx
Ja
=-
J
1
CLinearity)
Let
A
,
B
be
constants
,
then
((A fix
+
Bgxs)dx
=
Al
fixdx+
BS
!
gixdx
ids
!
fix
dx+)
;
fixdx
=
Sc
fix
&x
(e)
If
a
b
and
fixe
<
gix
s
for
as
X2b
,
them
(fixsax
?
gix,dx
1)
Triangle
inequality)
((fixdx)</
-
/1x >1dx
symmetric
proper ties
191
Let
aso
and
fix
s
be
n
on
<-a
,
as
[
then
1 fixdx
=
0
·
fl
-
x)
=
-
f(x)
,
fl0)
=
0
.
14
>
Let
aso
and
fix
s
be
tion
on
S-a
,
as
then
6
+1x
Ax
=
2)"
fixdx
·
f(
-
x1
=
fix)
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Example
11
(
(x+1)dx
=
So
xdx
+
J
!
1dx
=
+
3
MJ3
xdx
=
(0xdx
+
13xdx
y
=
Soxdx
+
1
.
!
+
d
+
=
0
en
write
X
=
-
t
I
X
since
-
runs
from
-3
to
0
then
X
runs
from
3
to
0
(
-
%
+
dt
=
fixx)d(
x)
=
+y
,
0xdx
=
-
!
xdx
1)
S
!*
&x
no
meaning
The
integration
is
improper
since
the
function
*
has
no
definition
for
x
=
0
.
fixs
=
*
is
an
odd
function
on
Df=
/R1/03
delete
origin
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Theorem
(mean-value
Theorem
for
integrals)
Let
f
be
continuous
on
a
closed
inter val
[a
.
b]
,
then
there
exist
a
point
(-[9
,
33
such
that
(fixdx
=
(b
-
a) f(x)
.
Proof
Recall
the
Inter mediate
-
Value
Theorem
:
If
gix
is
continuous
on
Sa
,
b]
,
then
it
takes
all
value s
between
its
minimal
value
and
its
maximum
value
·
Now
&
obtains
its
minimal
value
m
(at
x
=
1)
and
its
maximum
value
in
(at
x
=
U
,
i
.
e
.
M
=
fine)
.
m(b-
a)
<
(af(x)
ax
-
>
M(b
-
a)
i
.
e
.
fiel
=
Safatix dx<
f(u)
.
By
the
Intermediate
-
value
Theorem
,
fixe
must
take
on
every
value
between
fle)
and
fius
.
So
there
is
some
point
between
1
and
a
such
that
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fxx
=
(a
fix)dx
-
we
call
it
the
-
mean
value
of
we
rewrite
it
as
f
on
Ca
.
b3
.
Safixdx
=
(b
-
a)
f(C)
.
I
Definite
integrals
of
piecewise
continuous
functions
.
Let
Co<C
,
<
...
<Cn
on
IR
A
function
f
defined
on
[Co
,
Cn]
except
possibly
at
some
of
those
Ci
,
is
called
piecewise
continuous
if
for
each
:
(1
:
<n)
there
exist s
a
continuous
function
Fi
on
the
closed
set
Sit
,
(i)
such
that
f(x)=
F
:
on
open
set
(Cit
.
Ci)
.
Now
I"
fix
&x
t
I
.
Fixedx
Y
=
I
is
NOT
piecewis
continuous
on
St
.
12
.
